Aufgaben:Exercise 3.10Z: Maximum Likelihood Decoding of Convolutional Codes

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Overall system model,  given for this exercise

The Viterbi algorithm represents the best known realization form for the maximum likelihood decoding of a convolutional code.  We assume here the following model:

• The information sequence  $\underline{u}$  is converted into the code sequence  $\underline{x}$  by a convolutional code.

• It is valid  $u_i ∈ \{0, \, 1\}$.  In contrast,  the code symbols are represented bipolar   ⇒   $x_i ∈ \{–1, \, +1\}$.

• Let the channel be given by the models  $\text{BSC}$    ⇒   received values  $y_i ∈ \{–1, \, +1\}$  or  $\text{AWGN}$   ⇒ $y_i$  real-valued.

• Given a received sequence  $\underline{y}$  the Viterbi algorithm decides for the sequence  $\underline{z}$  according to

$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.03cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} |\hspace{0.05cm} \underline{y} ) \hspace{0.05cm}.$$

• This corresponds to the  "Maximum a posteriori"  $\rm (MAP)$  criterion.  If all information sequences  $\underline{u}$  are equally likely,  this transitions to the somewhat simpler  "Maximum likelihood criterion"  $\rm (ML)$:

$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{y} \hspace{0.05cm}|\hspace{0.05cm} \underline{x}_{\hspace{0.03cm}i} ) \hspace{0.05cm}.$$

• As a further result,  the Viterbi algorithm additionally outputs the sequence  $\underline{v}$  as an estimate for the information sequence  $\underline{u}$.

In this exercise,  you should determinethe relationship between the  $\text{Hamming distance}$  $d_{\rm H}(\underline{x}, \, \underline{y})$  and the  $\text{Euclidean distance}$

$$d_{\rm E}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) = \sqrt{\sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2}\hspace{0.05cm}.$$

Then,  the above maximum likelihood criterion is to be formulated with

• the Hamming distance  $d_{\rm H}(\underline{x}, \, \underline{y})$,

• the Euclidean distance  $d_{\rm E}(\underline{x}, \, \underline{y})$,  and

• the  $\text{correlation value}$  $〈 x \cdot y 〉$.

Hints:

• This exercise refers to the chapter  "Decoding of Convolutional Codes".

• Reference is made in particular to the section  "Viterbi algorithm – based on correlation and metrics".

• For simplicity,  "tilde"  and  "apostrophe"  are omitted.

• For more information on this topic,  see the following sections in this book:

• "MAP and ML criterion",

• "ML decision at the BSC channel",

• "ML decision at the AWGN channel",

• "Decoding linear block codes".

Questions

1

How are   $d_{\rm H}(\underline{x}, \, \underline{y})$   and   $d_{\rm E}(\underline{x}, \, \underline{y})$   related in the BSC model?

	$d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}(\underline{x}, \, \underline{y})$  is valid.
	$d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}^2(\underline{x}, \, \underline{y})$  is valid.
	$d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}^2(\underline{x}, \, \underline{y})/4$  is valid.

2

Which of the equations describe the maximum likelihood decoding in the BSC model?  The minimization/maximization refers alwaysto all  $\underline{x} ∈\mathcal{ C}$.

	$\underline{z} = \arg \min {d_{\rm H}(\underline{x}, \, \underline{y})}$,
	$\underline{z} = \arg \min {d_{\rm E}(\underline{x}, \, \underline{y})}$,
	$\underline{z} = \arg \min {d_{\rm E}^2(\underline{x}, \, \underline{y})}$,

3

Which equation describes the maximum likelihood decision in the BSC model?

	$\underline{z} = \arg \min 〈 \underline{x} \cdot \underline{y} 〉$,
	$\underline{z} = \arg \max 〈 \underline{x} \cdot \underline{y} 〉$.

4

What equations apply to the maximum likelihood decision in the AWGN model?

	$\underline{z} = \arg \min {d_{\rm H}(\underline{x}, \, \underline{y})}$,
	$\underline{z} = \arg \min {d_{\rm E}(\underline{x}, \, \underline{y})}$,
	$\underline{z} = \arg \max 〈 \underline{x} \cdot \underline{y} 〉$.

Solution

Solution

(1)  Correct is the  proposed solution 3:

• Let the two binary sequences be   $\underline{x}$   and   $\underline{y}$   with   $x_i ∈ \{-1, \, +1\}, \ y_i ∈ \{-1, \, +1\}$.  Let the sequence length be  $L$  in each case.

• The Hamming distance   $d_{\rm H}(\underline{x}, \, \underline{y})$   gives the number of bits in which  $\underline{x}$  and  $\underline{y}$  differ,  for which thus  $x_i \, - y_i = ±2$   ⇒   $(x_i \, - y_i)^2 = 4$  holds.

• Equal symbols  $(x_i = y_i)$  do not contribute to the Hamming distance and give  $(x_i \, – y_i)^2 = 0$.  According to the  solution 3,  we can therefore write:

$$d_{\rm H}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) = \frac{1}{4} \cdot \sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2= \frac{1}{4} \cdot d_{\rm E}^2(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.05cm}.$$

(2)  All proposed solutions  are correct:

• In the BSC model,  it is common practice to select the code word  $\underline{x}$  with the smallest Hamming distance  $d_{\rm H}(\underline{x}, \, \underline{y})$  for the given received vector  $\underline{y}$:

$$\underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm H}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.05cm}.$$

• But according to the subtask  (1)  also applies:

$$\underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})/4 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm E}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) \hspace{0.05cm}.$$

• The factor  $1/4$  does not matter for the minimization.  Since  $d_{\rm E}(\underline{x}, \, \underline{y}) ≥ 0$,  it does not matter whether the minimization is done with respect to   $d_{\rm E}(\underline{x}, \, \underline{y})$   or   $d_{\rm E}^2(\underline{x}, \, \underline{y})$.

(3)  Correct is the  proposed solution 2:

• The square of the Euclidean distance can be expressed as follows:

$$d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) = \sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2 = \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2} \hspace{0.1cm}-2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot y_i \hspace{0.05cm}.$$

• The first two summands are each equal to  $L$  and need not be considered for minimization.

• For the last expression in this equation,  $–2 \cdot 〈 \underline{x}, \, \underline{y} 〉$  can be written.

• Due to the negative sign,  minimization becomes maximization   ⇒   answer 2.

(4)  Correct are the  proposed solutions 2 and 3:

• For the AWGN channel,  unlike the BSC,  no Hamming distance can be specified.

• Based on the equation

$$d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) = \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2} \hspace{0.1cm}-2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot y_i$$

the same statements apply for the first and last summands as for the BSC model – see subtask  (3).

• For the middle summand,  $y_i = x_i + n_i$  and  $x_i ∈ \{–1, \, +1\}$  hold:

$$\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2} = \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} n_i^{\hspace{0.15cm}2} \hspace{0.1cm}+2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot n_i \hspace{0.05cm}.$$

• The first summand gives again  $L$,  the second is proportional to the noise power,  and the last term vanishes since  $\underline{x}$  and  $\underline{n}$  are uncorrelated.

• So for minimizing  $d_{\rm E}(\underline{x}, \, \underline{y})$,  the sum over  $y_i^2$  need not be considered since there is no relation to the code sequences  $\underline{x}$.